Q.Solve when
The inequality simplifies to . For integer , the solution set is ; for real , it is the open interval .
The core idea here is that solving a linear inequality is almost identical to solving a linear equation — you isolate the variable using the same operations (adding, subtracting, multiplying, dividing). The only twist is that if you multiply or divide by a negative number, the inequality sign flips. That’s the one rule that catches students off guard.
In this problem, no sign-flip is needed, so the algebra is straightforward. The interesting part comes after: the solution set looks very different depending on whether is an integer or a real number. That’s the point of the question — to show that the same inequality can have different “answers” depending on the domain of .
Let’s work through it.
- Simplify the inequality. Start with . Subtract from both sides:
- Isolate the term with . Add to both sides:
- Divide by the coefficient of . Since , the inequality sign stays the same:
The solution to the inequality is .
Now we interpret this result for the two cases.
(i) is an integer.
Integers are whole numbers (…, -3, -2, -1, 0, 1, 2, 3, …). The condition means can be any integer strictly less than 2. That includes all negative integers, zero, and 1. It does not include 2 itself, because the inequality is strict ( not ).
So the integer solution set is:
Often in exams, they want this written in set-builder form or as a list. Since the set is infinite on the left, you can write:
A common mistake is to include 2 in the solution. Remember: means 2 is not a solution. Also, don’t forget negative integers — they satisfy the inequality too.
(ii) is a real number.
Real numbers include everything on the number line: integers, fractions, decimals, irrationals. The condition now describes all real numbers to the left of 2, but not including 2 itself. This is the open interval from negative infinity to 2.
In interval notation:
In set-builder notation:
When the domain is “real numbers,” the solution is almost always an interval. When the domain is “integers,” you list the integer values that satisfy the condition. Always check the domain before writing your final answer.
For integer , the solution is ; for real , the solution is .
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